Next, suppose that \(f_j\)  are convex. This means that all \(\nabla ^2f_j()\)  are positive semidefinite, and so is \({\hat{}}()\) . Thus we can use the matrix Chernoff inequality [1]}, which provides a sharper guarantee:
\(\mathbb {P}\big (\Vert {\hat{}}()-()\Vert \ge \eta \big )&\le 2d\exp \bigg (-c\min \Big \lbrace \frac{\eta ^2s}{\lambda _{\min }\lambda _{\max }R},\frac{\eta s}{\lambda _{\max }R}\Big \rbrace \bigg ) \\&\le 2\exp \bigg (-c^{\prime }\min \Big \lbrace \frac{\eta ^2s}{\lambda _{\min }\lambda _{\max }R\log (d)},\frac{\eta s}{\lambda _{\max }R\log (d)}\Big \rbrace \bigg )\\&\le 2\exp \bigg (-c^{\prime \prime }\min \Big \lbrace \frac{\eta }{\sqrt{\lambda _{\min }\lambda _{\max }R\log (d)/s}},\frac{\eta }{\lambda _{\max }R\log (d)/s}\Big \rbrace \bigg )\\&\le 2\exp \bigg (-\frac{c^{\prime \prime }\eta }{\sqrt{\lambda _{\min }\lambda _{\max }R\log (d)/s} + \lambda _{\max }R\log (d)/s}\bigg ).\) 
